document.write( "Question 52282: If the sides of a square are decreased by 3cm, the area is decreased by 81cm squared. What were the dimensions of the original square? I was also wondering if you could show the work you used to get to the solution, thank you. \n" ); document.write( "

Algebra.Com's Answer #34837 by Earlsdon(6294) $\"\"$ $\"About$

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Let the original side of the square be s.
\n" ); document.write( "The area of the original square is: $\"A+=+s%5E2\"$
\n" ); document.write( "If you decrease s by 3 you'll have s-3.
\n" ); document.write( "The area of the new square will be:
\n" ); document.write( " $\"a+=+%28s-3%29%5E2\"$
\n" ); document.write( " $\"a+=+s%5E2-6s%2B8\"$ and this is equal to the original area $\"s%5E2\"$ decreased by 81. Putting this all together, we get:
\n" ); document.write( " $\"s%5E2-6s%2B9+=+s%5E2-81\"$ Simplify and solve for s, the length of the side of the original square. Subtract $\"s%5E2\"$ from both sides.
\n" ); document.write( " $\"-6s%2B9+=+-81\"$ Subtract 9 from both sides.
\n" ); document.write( " $\"-6s+=+-90\"$ Divide both sides by -6.
\n" ); document.write( " $\"s+=+15\"$ \r
\n" ); document.write( "\n" ); document.write( "The dimensions of the original square were:
\n" ); document.write( "15 cm. by 15 cm. \n" ); document.write( "

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